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Math Help - squares

  1. #1
    Member Veronica1999's Avatar
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    Jupimar
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    squares

    I can't seem to figure out the logic behind this problem.
    Would this problem have something to do with mode?
    I know I am missing the main point of this problem.
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  2. #2
    Senior Member
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    Nov 2011
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    Crna Gora
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    Re: squares

    Since 2^{n-12} +49 is an odd number it must be perfect square also . So:

    x^2=2^{n-12}+49 \Rightarrow 2^{n-12}=x^2-49

    Now, if we make substitution y=n-12 we can write :

    2^y=x^2-49 \Rightarrow y= \log_{2} (x^2-49)

    This equality has integer solutions only if x^2-49 is power of 2 so x must be an odd number so we shall make substitution x=2m-1 and therefore :

    y=\log_{2} (4m^2-4m-48)=\log_{2}(4 \cdot(m^2-m-12))

    =2+\log_{2} (m^2-4m+3m-12)=

    =2+\log_{2} (m+3)(m-4)=2+\log_{2} (m+3)+\log_{2}(m-4)

    It is obvious that both m+3 and m-4 must be powers of 2 and this is possible only if m=5 , therefore :

    y=2+3+0 \Rightarrow y=5 \Rightarrow n-12=5 \Rightarrow n=17 .
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